[LeetCode]Add Digits

Given a non-negative integer num, repeatedly add all its digits until the result has only one digit.

For example:

Given num = 38, the process is like: 3 + 8 = 11, 1 + 1 = 2. Since 2 has only one digit, return it.

利用數字根的規律求解（Digital Root）

一個數的數字根是它除以9的餘數（如果該數是9的倍數，那麼該數的數字根就是9）

歸納爲：dr(n)=1+(num-1)%9

public class Solution {
   public int addDigits(int num) {
        return 1 + (num - 1) % 9;
    }
}